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CIE 9231 2024 June Paper 21 Q6

A Level / CIE / FP2

CIE 9231 2024 June Paper 21 Paper · Question 6

题目

Problem

(a) Show that

(coshx+sinhx)12=e12x.\left(\cosh x+\sinh x\right)^{\frac{1}{2}}=e^{\frac{1}{2}x}.
[2]

(b) Find the particular solution of the differential equation

d2ydx2+dydx+3y=5(coshx+sinhx)12,\frac{\mathrm{d}^2y}{\mathrm{d}x^2}+\frac{\mathrm{d}y}{\mathrm{d}x}+3y=5\left(\cosh x+\sinh x\right)^{\frac{1}{2}},

given that, when x=0x=0, y=1y=1 and dydx=43\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{4}{3}.

[10]
题目中文翻译

(a) 证明

(coshx+sinhx)12=e12x\left(\cosh x+\sinh x\right)^{\frac{1}{2}}=e^{\frac{1}{2}x}

(b) 求微分方程

d2ydx2+dydx+3y=5(coshx+sinhx)12\frac{\mathrm{d}^2y}{\mathrm{d}x^2}+\frac{\mathrm{d}y}{\mathrm{d}x}+3y=5\left(\cosh x+\sinh x\right)^{\frac{1}{2}}

的特解,已知当 x=0x=0 时,y=1y=1dydx=43\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{4}{3}

解答